Problem: Perform the row operation, $3R_1\rightarrow R_1$, on the following matrix. $\left[\begin{array} {ccc} 2 & 1 & 2 & 0 \\ -4 & 4 & -7 & -1 \\ 0 & 9 & 1 & 9 \end{array} \right] $
Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_1$ is given below. $R_1=\left[\begin{array} {ccc} 2 & 1 & 2 & 0 \end{array} \right]$ We are asked to perform the row operation, $3R_1\rightarrow R_1$. Therefore, we must multiply $R_1$ by $3$. $\begin{aligned}3R_1 &=3\left[\begin{array} {ccc} 2 & 1 & 2 & 0 \end{array} \right] \\\\&=\left[\begin{array} {ccc} 6 & 3 & 6 & 0 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_1$ with $3R_1$. $\left[\begin{array} {ccc} {2} & {1} & {2} & {0} \\ -4 & 4 & -7 & -1 \\ 0 & 9 & 1 & 9 \end{array} \right]\xrightarrow{3R_1\rightarrow R_1} \left[\begin{array} {ccc} {6} & {3} & {6} & {0} \\ -4 & 4 & -7 & -1 \\ 0 & 9 & 1 & 9 \end{array} \right]$ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} 6 & 3 & 6 & 0 \\ -4 & 4 & -7 & -1 \\ 0 & 9 & 1 & 9 \end{array} \right]$